Solving A Quintic Without Using The Quintic Formula

[@manishkumargond697]

7:17 it should be a(e-b) not a(e-a)

[@NoonMemeWow]

I was looking for this comment. I was unable to concentrate on the rest of the video knowing that a mistake had been made, and it was not corrected. I am not sure if the final values for a,b,c,d,e have been computed correctly

[@holyshit922]

Golden ratio and number opposite to reciprocal of golden ratio are the roots This allows to reduce problem to solving cubic with integer coefficients

[@rainerzufall42]

You probably mean φ and - 1/φ. Alternatively φ and 1 - φ.

[@rainerzufall42]

10:47 Weak! Those number don't even fit the last equation! With a = 1, e = -1, you get: a²e - ae² = -1 - 1*1 = -2, not -3! Okay, that's because the formula was wrong: -3/e = a (b - e) = a (a² - e - e) = a (a² - 2e) = a³ - 2ae => ae (a² - 2e) = a³e - 2ae² = -3 || 1³(-1) - 2*1*1 = -3 ✅

[@rainerzufall42]

We didn't even determine a, b, c, d, e, you just gave it to us. Not even speaking about solving the cubic equation...

[@rainerzufall42]

7:17 a (e - b) ... => c = a (b - e)

[@NadiehFan]

The key to solving the equation x⁵ = 5x + 3 is to note that for x = φ (golden ratio) and x = ψ = −1/φ and any positive integer n we have xⁿ = Fₙx + Fₙ₋₁ where Fₙ is the n-th Fibonacci number. So, it is immediately obvious that the two roots φ and ψ of x² = x + 1 are roots of x⁵ = 5x + 3, meaning that x² − x − 1 is a factor of x⁵ − 5x − 3. @SyberMath: if you want a _really_ challenging (but algebraically solvable) quintic equation you should try x⁵ − 5x³ + 5x = 1 or x⁵ + 5x³ + 5x = 1 These equations look very similar, but the strategies for solving them are different. Can you do it?

[@holyshit922]

Problem Find condition which must be satisfied to be able to decompose quintic into (x^3+ax^2+bx+c)(x^2+dx+e)=x^5+a_{4}x^4+a_{3}x^3+a_{2}x^{2}+a_{1}x+a_{0} What codition coefficients of quintic must satisfy to make this decomposition possible Coefficients of this quintic satisfies that condition

[@stevenlitvintchouk3131]

With your equations and the ones folks post on X, it's gotten to the point that one of the first things I do is to guess that maybe the golden ratio is involved somewhere. And in this case, the given polynomial is indeed divisible by (x^2 - x - 1), which is how I cracked this problem.

[@roberttelarket4934]

It’s excellent to show such type quintics to the uniniated!

[@SyberMath]

I'm glad you found it helpful!

[@roberttelarket4934]

@@SyberMath: Uninitiated was incorrectly spelled and if I edited now I would lose the ❤️. Also I made the statement for others. I'm aware of the difficulty with quintics.

[@arekkrolak6320]

Solving equation by guessing solution :)

[@yuryp6975]

Fibinacci dequence gives the 2 roots (golden ratio snd the reciprocal)

[@Christopher-e7o]

X,2×+5=8

[@RayMyName]

i actually guessed the solution by seeing that one must be between 1 and 2 so the golden ratio was the first thing to come to mind after trying some ratios of 5

[@philipfoy7117]

I did the same

[@pwmiles56]

@@philipfoy7117 Me too. You can verify by plugging in x^2=x+1 and doing repeated substitutions. As a bonus, the negative root i.e. (1-sqrt(5))/2 must also be a solution.

[@stephenshefsky5201]

When you see coefficients that are Fibonacci numbers, as are 3 and 5, think about the Golden Ratio!

[@pmccarthy001]

You mean it's impossible to resolve the complex relationships between the roots and coefficients of all (most) polynomial equations of one variable where the degree of the polynomial is 5, or above... with the operations of addition, subtraction, multiplication, division, and radicals. We do know of operations, like theta functions, that can resolve the relationships between the roots and coefficients of some of this class of polynomial equations which cannot be solved by simple arithmetic operations and the taking of roots. However, I've never seen any evidence that anyone's discovered, or invented, operations that can resolve the relationships between the roots and coefficients of any polynomial equation of one variable of any degree. But I don't believe anyone's ever proven that such operations that might be able to do this can not exist. I believe it also might be possible to reconceptionalize the problem to see polynomials of one variable in some light that might suggest how to resolve these relationships. Perhaps different number systems, some new conceptionalization of what a univariate polynomial is? But with the tools and conceptonalizations we've come up with so far, no one's been able to come up with such operations. I don't know, but where mathematics often seems to deal with abstractions upon abstractions it's difficult to predict what someone might one day conceive of.

[@sjn7220]

This is why we have Newton-Raphson 😀

[@giuseppemalaguti435]

La funzione f(x)=x^5-5x-3 è negativa per x=0,positiva per x=-1...quindi uno zero sarà compreso tra )-1,0( e,casualmente, può essere x=(1-√5)/2...

[@SweetSorrow777]

How did you know that the factors were a quadratic and cubic instead of a quartic and linear? Guessing?

[@SyberMath]

That's how the problem was made up but I think this is easier in general. If you had a quartic that would be factored into two quadratics and one of them could be multiplied by the linear factor to produce a cubic. I could be wrong, though since I haven't checked it

[@scottleung9587]

Nice job!

[@SyberMath]

Thank you! 😍

[@promixinc.8434]

Finally, I have a one good reason to learn newton-raphson. Great video